Aus Online Mathematik Brückenkurs 1

We can factorize the denominators as

\displaystyle y^{2}-2y=y\left( y-2 \right)

\displaystyle y^{2}-4=\left( y-2 \right)\left( y+2 \right) [by the conjugate rule]

and then we see that the terms' lowest common denominator is \displaystyle y\left( y-2 \right)\left( y+2 \right) because it is the product that contains the smallest number of factors which contain both \displaystyle y\left( y-2 \right) and \displaystyle \left( y-2 \right)\left( y+2 \right) .

Now, we rewrite the fractions so that they have same denominators and start simplifying:

\displaystyle \begin{align} & y^{2}-4=\left( y-2 \right)\left( y+2 \right) \\ & y^{2}-2y=y\left( y-2 \right) \\ & \frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}=\frac{1}{y\left( y-2 \right)}\centerdot \frac{y+2}{y+2}-\frac{2}{\left( y-2 \right)\left( y+2 \right)}\centerdot \frac{y}{y} \\ & \\ & =\frac{y+2}{y\left( y-2 \right)\left( y+2 \right)}-\frac{2y}{\left( y-2 \right)\left( y+2 \right)y} \\ & \\ & =\frac{y+2-2y}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)} \\ \end{align}

The numerator can be rewritten as \displaystyle -y+2=-\left( y-2 \right) and we can eliminate the common factor \displaystyle y-2 .

\displaystyle \frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-\left( y-2 \right)}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-1}{y\left( y+2 \right)}=-\frac{1}{y\left( y+2 \right)}